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Answer to Question #4189 in Calculus for shane duque

Question #4189
A man 6 ft. tall walks away from a lamppost 16 ft. high at a rate of 5 mi per hr. How fast does the of his shadow move?
Suppose a man was in point A and the shadow was in point S1 at time t = t0. After some time t, the man came to the point B and the shadow – to the point S2. OL is a lamppost.
As we want to evaluate the speed of the shadow, we need to find:
(S1 S0)/t = (S1 B+BS2)/t Angles <OS1L and <CS1A are equal, so:
OL/(LA+AS1 )=AC/(AS1 )
AS1*H = AS1*h + LA*h
Where H is the height of the lamppost (H = 16ft) and h is the man’s height (h = 6ft).
AS1 = LA*h/(H-h)
So, S1 B = AB - AS1 = Vt - LA*h/(H-h)
Angles <OS2 L and <DS2 B are equal, so:
OL/(LA+AB+BS2 ) = BD/(BS2 )
BS2*H = LA*h + Vt*h + BS2*h
BS2 = h(LA+Vt)/(H-h)
Then:
(S1 S2)/t = (S1 B+BS2)/t = V - LA/t*h/(H - h) + h(LA + Vt)/(H - h)t = V + (h*LA + Vth - LA*h)/((H - h)t)
= V + V*h/(H - h) = VH/(H - h)
The speed of the shadow is 5*16/(16-6)=8 miles/hour

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